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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Show that $\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
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Answer

We have,

$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$\frac{\text{dy}}{\text{dx}}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(2)$

Differentiating both sides of equation (2) with respect to 3, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}+2\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\big(2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}\big)+2\big(\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}\big)$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$

Hence, the given function is the solution to the given differential equation.

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